I am having a bit of trouble figuring out how exactly to make a certain connection. It's a project I thought might be simple enough for me to do on my own without help, but I've hit a wall unfortunately. It's an 'exercise generator' that basically asks a few basic questions, and based on your answers, it outputs a recommended workout routine. I've constructed the MySQL database with plenty of exercises, have successfully connected the database, and have made the form.
What I am having trouble though, however, is storing those form results into variables, and based on those variables (if workout days is 3, for example, there would only be 3 groups of workouts printed instead of 5) output a routine into the same div as the form, effectively replacing it with the answer instead of placing it underneath the submitted form.
Index.php
<!DOCTYPE html>
<html>
<?php $page_title = "Workout Generator"; ?>
<link rel="stylesheet" type="text/css" href="style.css">
<head>
<script type="text/JavaScript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js" ></script>
</head>
<body>
<?php include("header.php"); ?>
<?php include("connect.php"); ?>
<div id="appWindow">
<h3>Please answer the following questions and click 'Submit' to generate your workout routine.</h3>
<form id="homeForm" method="post" >
<label for="name">Name: </label>
<input type="text" name="name"><br>
<label for="age">Age: </label>
<input type="number" name="age"><br>
<label for="workoutdays">How many days a week can you workout?</label>
<select name="workoutdays">>
<option value="1">3</option>
<option value="2">5</option>
</select><br>
<label for="workoutstyle">Do you prefer a gym, or bodyweight exercises?</label>
<select name="workoutstyle">>
<option value="1">Gym</option>
<option value="2">Bodyweight</option>
</select><br>
<button type="submit" name="submit">Submit</button>
<button type="reset" name="reset">Reset</button>
<div class="form_result"></div>
</form>
<?php
$age = $_POST['age'];
$workoutdays = $_POST['workoutdays'];
$workoutstyle = $_POST['workoutstyle'];
?>
</div>
<br><br><br>
<?php include("footer.php"); ?>
</body>
</html>
I don't necessarily want an answer giving me the exact code to enter, but would appreciate being pointed in the right direction to get that form pulling data from MySQL, and using AJAX to print in the same window without refreshing.
THANK YOU
In the script file
$(document).ready(function(){
$(document).on('click','#submit_btn',function(){
var url = '/xyz.php'; //full path of the function where you have written the db queries.
var data = '';//any data that you would like to send to the function.
$.post(url,{data: data}, function(result){
var arr = JSON.parse(result);
});
});
});
in your php file once your database query has been executed and you get the result. for example
$result = //result of your mysql query.
echo json_encode($result);
exit;
First of all, you need to post your request. With $.ajax this
//Do this thing on page load
$(function() {
//handle submit
$("#homeForm").submit(function(e) {
//customize your submit
e.preventDefault();
$.ajax({
type: "POST",
url: youURL, //maybe an url pointing to index.php
data: yourData, //attach everything you want to pass
success: function(response) {
$("#appWindow").html(response);
}
});
});
});
code should help. You need to make sure you pass the necessary elements and you provide the correct url. On server side, generate the desired html and send back as response.